UVa 10277 boastin ' Red Socks (enum)

10277-boastin ' Red Socks

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=115&page=show_ problem&problem=1218

You have a drawer this is all two kinds of socks:red and black. You are know that there are in least 2 socks, and not more than 50000. However, you don't know how many there actually are,

Nor do your know how many are red, or to many are black. (Your mother does the laundry!)

You have noticed, though, which is the drawer each morning and choose two to socks (in wear pitch Cannot distinguish red from black), the probability

Pick two red socks is exactlyp/Q, where 0 < Q and 0 <= p <= Q.

From this, can your determine how many socks of each colour are in your drawer? There May is multiple solutions-if so pick the solution with the fewest total number of socks, but still allowing

You have to wear a couple of same color socks.

Input

Input consists of multiple problems, each on a separate line. Each problem consists of the integersp andQ separated by a. Note thatP and Q would both fit into a unsigned long integer.

The Input is terminated by a line consisting of two zeroes.

Output

For each problem, output a single line consisting of the number of red socks and the number of black socks in your drawer, separated by one space. If There is no solution to the problem, print "impossible".

Sample Input

See more highlights of this column: http://www.bianceng.cn/Programming/sjjg/

1 2
6 8
2499550020
789
0 0

Sample Output

3 1
7 1
4 49992
Impossible

Train of thought: Set red socks have x only, black sock has y only, then have

C (x,2)/C (x+y,2) =x (x-1)/((X+y) (x+y-1)) =p/q

So enumerate the denominator I to see if it satisfies q| (i* (I-1)), and then see if the corresponding p can be split into j* (j-1) Form

PS: This problem can be translated into a two-yuan two-time Diophantine equation

[(Q-p) (2x-1)-P (2y)]^2-PQ (2y) ^2 = (q-p) ^2, when the q-p!=1 is not necessarily a solution, more specific analysis? Pit

Full code: (Run 0ms because of a UVA server failure at submission time (commonly known as Aunt))

 01./*0.000s*/02.  
#include <cstdio> #include <cmath> 05.typedef unsigned int ll;  
06.const ll maxn = 50000; 08.ll gcd (ll A, ll b) 09.  
{A. return b gcd (b, a% B): A;  
11.} 12. 13.int Main () 14.  
{ll p, q, G, I, pp;  
Double J;    while (scanf ("%u%u", &p, &q), Q) 18.        {. if (P = = 0)///probability is 0 20.            {puts ("0 2");///at least 2 socks 22.  
Continue        23.} 24.  
g = gcd (p, q);  
P/= G, Q/= g;            for (i = 2; I <= MAXN, ++i)///not more than 50000 27.            if (i * (i-1)% q = = 0)///The denominator 28 of the condition.  
{pp = i * (i-1)/q * p;  
j = sqrt (0.25 + pp) + 0.5;                if (round (j) = j) The///molecule also meets the condition 32.  
{A. printf ("%u%u\n", (LL) J, I-(LL) j);  
break;       35.} 36.     } 37.  
if (i > Maxn) puts ("impossible");    38.} 39.  
return 0; 
.}